Solving Trig Equations

• Nov 24th 2007, 01:57 PM
Macleef
Solving Trig Equations
Solve each equation for $\displaystyle 0 \leq \theta \leq 360$

--
$\displaystyle sin^2x - 1 = cos^2x$
--

Work:
$\displaystyle cos^2x - cos^2x = 0$
$\displaystyle 0 = 0$

--
90 and 270
--

Incorporate the use of the double angle formulas identities and other identities to solve the following on $\displaystyle 0 \leq \theta \leq 360$

--
$\displaystyle sin2x = 2cos2x$
--

Work:
$\displaystyle 2sinxcosx - 2cosx = 0$
$\displaystyle 2cosx (sinx - 1) = 0$

What do I do now?

--
90 and 270
--

Btw, how would you simply $\displaystyle cos^2x$ to $\displaystyle cosx$ if you had something like $\displaystyle cos^2x = 1$? Would it be...
$\displaystyle cosx = \sqrt{1}$
$\displaystyle x = cos^-1 \sqrt{1}$?
• Nov 24th 2007, 02:10 PM
topsquark
Quote:

Originally Posted by Macleef
--
$\displaystyle sin^2x - 1 = cos^2x$
--

Work:
$\displaystyle cos^2x - cos^2x = 0$
$\displaystyle 0 = 0$

Not quite. $\displaystyle 1 - sin^2(x) = cos^2(x)$, so
$\displaystyle sin^2x - 1 = cos^2x$

$\displaystyle -cos^2(x) = cos^2(x)$

$\displaystyle 2cos^2(x) = 0$

$\displaystyle cos(x) = 0$

Thus $\displaystyle x = 90^o , 270^o$

-Dan
• Nov 24th 2007, 02:23 PM
topsquark
Quote:

Originally Posted by Macleef
Incorporate the use of the double angle formulas identities and other identities to solve the following on $\displaystyle 0 \leq \theta \leq 360$

--
$\displaystyle sin2x = 2cos2x$
--

Work:
$\displaystyle 2sinxcosx - 2cosx = 0$
$\displaystyle 2cosx (sinx - 1) = 0$

What do I do now?

This is one of the nastier ones:
$\displaystyle sin(2x) = 2cos(2x)$

$\displaystyle 2sin(x)cos(x) = 2cos(2x)$

$\displaystyle sin(x) cos(x) = cos(2x)$

$\displaystyle sin(x) cos(x) = 2cos^2(x) - 1$

$\displaystyle cos(x) (\pm)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1$

$\displaystyle cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1$

$\displaystyle 5cos^4(x) - 5cos^2(x) + 1 = 0$

This has solutions
$\displaystyle cos(x) = \pm 0.85065080835215 \implies x = 31.7175^o, 148.283^0$
and
$\displaystyle cos(x) = \pm 0.52573111211905 \implies x = 58.2825^o, 121.717^o$

Of these, only $\displaystyle x = 31.7175^o, 121.717^o$ solve the original equation.

-Dan
• Nov 24th 2007, 02:25 PM
topsquark
Quote:

Originally Posted by Macleef
Btw, how would you simply $\displaystyle cos^2x$ to $\displaystyle cosx$ if you had something like $\displaystyle cos^2x = 1$? Would it be...
$\displaystyle cosx = \sqrt{1}$
$\displaystyle x = cos^-1 \sqrt{1}$?

How do you solve any equation where you take the square root of both sides?

$\displaystyle cos^2(x) = 1$

$\displaystyle cos(x) = \pm 1$

Thus
$\displaystyle x = cos^{-1}(1) = 0^o$
and
$\displaystyle x = cos^{-1}(-1) = 180^o$

-Dan
• Nov 24th 2007, 02:59 PM
Macleef
Quote:

Originally Posted by topsquark

This is one of the nastier ones:
$\displaystyle sin(2x) = 2cos(2x)$

$\displaystyle 2sin(x)cos(x) = 2cos(2x)$

$\displaystyle sin(x) cos(x) = cos(2x)$

$\displaystyle sin(x) cos(x) = 2cos^2(x) - 1$

$\displaystyle cos(x) (\pm)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1$

$\displaystyle cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1$

$\displaystyle 5cos^4(x) - 5cos^2(x) + 1 = 0$

This has solutions
$\displaystyle cos(x) = \pm 0.85065080835215 \implies x = 31.7175^o, 148.283^0$
and
$\displaystyle cos(x) = \pm 0.52573111211905 \implies x = 58.2825^o, 121.717^o$

Of these, only $\displaystyle x = 31.7175^o, 121.717^o$ solve the original equation.

-Dan

What the heck...? I'm so confused
I'm pretty sure this equation isn't that complicated...
• Nov 24th 2007, 03:01 PM
Macleef
Quote:

Originally Posted by topsquark
Not quite. $\displaystyle 1 - sin^2(x) = cos^2(x)$, so
$\displaystyle sin^2x - 1 = cos^2x$

$\displaystyle -cos^2(x) = cos^2(x)$

$\displaystyle 2cos^2(x) = 0$

$\displaystyle cos(x) = 0$

Thus $\displaystyle x = 90^o , 270^o$

-Dan

How did you get $\displaystyle -cos^2x$? Isn't it $\displaystyle cos^2x$ for $\displaystyle sin^2x - 1$?
• Nov 25th 2007, 06:26 AM
topsquark
Quote:

Originally Posted by Macleef
What the heck...? I'm so confused
I'm pretty sure this equation isn't that complicated...

This equation is. Whatever your book (teacher, professor, whatever) intended for you to solve certainly wasn't given the listed solution set. There's a typo in your given equation, but I can't figure out what it was meant to be.

-Dan
• Nov 25th 2007, 06:27 AM
topsquark
Quote:

Originally Posted by Macleef
How did you get $\displaystyle -cos^2x$? Isn't it $\displaystyle cos^2x$ for $\displaystyle sin^2x - 1$?

$\displaystyle sin^2(x) + cos^2(x) = 1$

$\displaystyle sin^2(x) + cos^2(x) - 1 = 0$

$\displaystyle sin^2(x) - 1 = -cos^2(x)$

-Dan